Tubular-Neighborhood Theorems (proof question)

I have a question about part of the proof of the Tubular Neighborhood Theorem, which I believe reduces to a question about the $$\epsilon$$-Neighborhood Theorem. First some preliminaries for my question to be precise.

Let $$(M,g)$$ be a Riemannian manifold, and let $$A\subset M$$ be an embedded submanifold. Let $$\xi$$ be any spray on $$M$$, with associated exponential map $$\exp:\mathcal{O}\to M$$, where $$\mathcal{O}$$ is some open domain of the tangent bundle $$TM$$. Let $$i(A)$$ denote the zero section in $$TM$$ (thinking of $$i:A\to TM$$ as the inclusion). Letting $$NA$$ denote the normal bundle of $$A$$, it's fairly straightforward to show that $$d(\exp)_{i(a)}:T_{i(a)}(\mathcal{O}\cap NA)\to T_aM$$ is the identity, and hence $$\exp$$ is local diffeomorphism around points on $$i(A)$$.

Now, we let $$\hat{g}$$ denote the Sasaki-metric related to $$g$$ on $$TM$$. This metric induces the (topological) metric $$d$$ on $$TM$$, turning $$(TM,d)$$, and hence $$(NA,d)$$ into a metric space. Since $$\exp(i(a))=a$$ for each $$a\in A$$, a standard metric space argument can then be made to show that there exists an open neighborhood $$U\subseteq\mathcal{O}\cap NA$$ of $$i(A)$$ so that $$\left.\exp\right|_U:U\to V$$ is a diffeomorphism, where $$V\subseteq M$$ is some open neighborhood of $$A$$.

Finally, we assume $$A$$ is a compact submanifold. I'm having trouble constructing an $$\epsilon$$-neighborhood of $$i(A)$$ in $$U$$. Let $$U^\epsilon=\{v\in N_aA:|v|_g<\epsilon, a\in A\}\subset NA.$$ I believe my question stems from a poor understanding of Sasakian-distance. Indeed, since $$U$$ is open, for each $$a\in A$$, there exists $$\epsilon(a)>0$$ such that $$B(i(a),\epsilon(a)):=\{(x,v)\in NA:d(i(a),(x,v))<\epsilon(a)\}\subseteq U.$$ Is $$\epsilon$$ continuous? That seems unclear to me, but could be trivial. Furthermore, if we did have some minimum, say $$\epsilon_0$$, so that $$\mathcal{B}:=\bigcup_{a\in A}B(i(a),\epsilon_0)\subseteq U,$$ how does one show that containment of $$U^{\epsilon_0}\subseteq\mathcal{B}$$?

Since $$\mathcal{B}$$ is dealing with Sasakian-geodesic balls, it seems reasonable, that we should have some lower bound on the injectivity radius of the Sasakian exponential map (when restricted to the zero section $$i(A)$$ in $$NA$$), and so the distance can be seen as just the Sasakian length of horizontal lifts (which are then easily related to $$|\cdot|_g$$). I can't seem to get my thinking precise here to make much sense of it however. This could also not be the best direction to take with the proof, so any help would be appreciated.

• I figured it out, though it wasn't exactly what I had in mind. I can construct a smooth $\epsilon:A\to\mathbb{R}$ via partition of unity argument and a locally-finite collection of charts which trivialize $A\times\{0\}$ (locally) that work. Which I suppose is the usual way, one goes about this proof. I'm a little concerned that there's no obvious way to do this the Sasakian metric however, so I'm going to leave this up, in case anyone can help my understanding of all things related to Sasakian distance.
– Matt
Sep 27 '18 at 8:15